Question

If f is a differentiable function, then the values $x=c$at which the sign of the derivative $f\text{'}\left(x\right)$ changes are the locations of the local extrema of f. Use this information to find the local extrema of the functionrole="math" localid="1649873816717" $f\left(x\right)=\sin x$. Illustrate your answer on a graph of$y=\sin x$.

Short answer

The local extrema of the given function is at $x=-1,0,1$. The graph of the given function is given below,

Step-by-step solution

Step 1. Given information.

Consider the given question,

The local extrema of the function $f\left(x\right)=\sin x$.

Step 2. Write the derivative of the given function.

The derivative of the given function,

$f\text{'}\left(x\right)=\cos x$

For local extrema, put $f\text{'}\left(x\right)=0$. Then,

role="math" localid="1649874004201" $$\begin{gathered} 0=\cos x \\ x=...,-\frac{3\mathrm{\pi }}{2},-\frac{\mathrm{\pi }}{2},0,\frac{\mathrm{\pi }}{2},\frac{3\mathrm{\pi }}{2},... \\ x=\frac{\left(2n+1\right)\mathrm{\pi }}{2} \end{gathered}$$

Where, n is an integer.

So,

$$\begin{gathered} f\left(\frac{\mathrm{\pi }}{2}\right)=\sin \left(\frac{\mathrm{\pi }}{2}\right) \\ =0 \\ f\left(-\frac{\mathrm{\pi }}{2}\right)=\sin \left(-\frac{\mathrm{\pi }}{2}\right) \\ =-1 \\ f\left(\frac{3\mathrm{\pi }}{2}\right)=\sin \left(\frac{3\mathrm{\pi }}{2}\right) \\ =-1 \\ f\left(-\frac{3\mathrm{\pi }}{2}\right)=\sin \left(-\frac{3\mathrm{\pi }}{2}\right) \\ =1 \end{gathered}$$

Therefore, the required graph is$y=\sin x$.

Step 3. Plot the function.

On plotting the function is given below,